The given expression \( e^{-\theta} \cos(6\theta) \, d\theta \) is an integral that can be solved using integration by parts. The formula for integration by parts is:

\[
\int u \, dv = uv - \int v \, du
\]

Let's solve the integral \( \int e^{-\theta} \cos(6\theta) \, d\theta \):

1. **Choose \( u \) and \( dv \):**
   - Let \( u = \cos(6\theta) \), so \( du = -6\sin(6\theta) \, d\theta \).
   - Let \( dv = e^{-\theta} \, d\theta \), so \( v = -e^{-\theta} \).

2. **Apply the integration by parts formula:**

\[
\int e^{-\theta} \cos(6\theta) \, d\theta = -e^{-\theta} \cos(6\theta) + 6\int e^{-\theta} \sin(6\theta) \, d\theta
\]

Now, we need to integrate \( \int e^{-\theta} \sin(6\theta) \, d\theta \). We can use integration by parts again for this integral.

3. **Choose \( u \) and \( dv \) for the new integral:**
   - Let \( u = \sin(6\theta) \), so \( du = 6\cos(6\theta) \, d\theta \).
   - Let \( dv = e^{-\theta} \, d\theta \), so \( v = -e^{-\theta} \).

4. **Apply integration by parts again:**

\[
\int e^{-\theta} \sin(6\theta) \, d\theta = -e^{-\theta} \sin(6\theta) - 6\int e^{-\theta} \cos(6\theta) \, d\theta
\]

Let's denote the original integral by \( I \), so:

\[
I = -e^{-\theta} \cos(6\theta) + 6\left[-e^{-\theta} \sin(6\theta) - 6I\right]
\]

5. **Solve for \( I \):**

\[
I = -e^{-\theta} \cos(6\theta) - 6e^{-\theta} \sin(6\theta) - 36I
\]

\[
37I = -e^{-\theta} \cos(6\theta) - 6e^{-\theta} \sin(6\theta)
\]

\[
I = \frac{-e^{-\theta}}{37} \left[\cos(6\theta) + 6\sin(6\theta)\right]
\]

So, the final answer is:

\[
\int e^{-\theta} \cos(6\theta) \, d\theta = \frac{-e^{-\theta}}{37} \left[\cos(6\theta) + 6\sin(6\theta)\right] + C
\]

where \( C \) is the constant of integration.

Would you like further details or have any questions? 

Here are 5 related questions:
1. How do you apply the method of integration by parts in other types of integrals?
2. Can integration by parts be used multiple times within a single integral?
3. What are the conditions under which integration by parts is most effective?
4. How does the choice of \( u \) and \( dv \) impact the complexity of solving an integral?
5. How can you solve integrals involving trigonometric functions using substitution?

**Tip:** When choosing \( u \) and \( dv \) in integration by parts, it's often helpful to pick \( u \) such that its derivative simplifies the expression, and \( dv \) as the part that can be easily integrated.

This page provides a detailed step-by-step solution to the integral of e^(-θ) * cos(6θ) using the integration by parts method. The problem involves key mathematical concepts such as integration, trigonometric functions, and exponential functions. Suitable for students in Grades 11-12.

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